What does this statement mean ? printf("[%.*s] ", (int) lengths[i],

Posted by Vivek Goel on Stack Overflow See other posts from Stack Overflow or by Vivek Goel
Published on 2011-03-14T08:07:06Z Indexed on 2011/03/14 8:10 UTC
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c

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regex

I was reading this page http://dev.mysql.com/doc/refman/5.0/en/mysql-fetch-row.html there is one line

printf("[%.*s] ", (int) lengths[i],
              row[i] ? row[i] : "NULL");

from code

    MYSQL_ROW row;
unsigned int num_fields;
unsigned int i;

num_fields = mysql_num_fields(result);
while ((row = mysql_fetch_row(result)))
{
   unsigned long *lengths;
   lengths = mysql_fetch_lengths(result);
   for(i = 0; i < num_fields; i++)
   {
       printf("[%.*s] ", (int) lengths[i],
              row[i] ? row[i] : "NULL");
   }
   printf("\n");

}

what does [%.*s] mean in that code ?

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What does this statement mean ? printf("[%.*s] ", (int) lengths[i], - Developer IT

What does this statement mean ? printf("[%.*s] ", (int) lengths[i],

c

regex

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